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8 months ago

14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards

1 answer:

8 0

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

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(1) A net force of 265 N accelerates a bike and rider at 2.30 m/s. What is the mass of the

kati45 [8]

Answer: 115.2kg

Explanation:

Net force = 265 N

Acceleration of bike & rider = 2.30m/s2 (The SI unit of acceleration is m/s2)

Mass of the bike and rider together = ?

Since force is the product of the mass of an object and the acceleration by which it moves, Force = Mass x Acceleration

265N = Mass x 2.30m/s2

Mass = (265N/2.30m/s2)

Mass = 115.2 kg

Thus, the Mass of the bike and rider together is 115.2kg

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3 years ago

A parallel-plate capacitor is formed from two 2.7 cm -diameter electrodes spaced 1.4 mm apart. The electric field strength insid

Andre45 [30]

Answer:

The potential difference between the plates is $8.4\times10^{3}\ V$

Explanation:

Given that,

Distance = 1.4 mm

Electric field strength $E= 6.0\times10^{6}\ N/C$

Let the potential difference is V.

We need to calculate the potential difference between the plates

Using formula of electric field

$E=\dfrac{V}{d}$

$V=Ed$

Where, V = potential

d = distance

Put the value into the formula

$V=6.0\times10^{6}\times1.4\times10^{-3}$

$V=8.4\times10^{3}\ V$

Hence, The potential difference between the plates is $8.4\times10^{3}\ V$

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2 years ago

Slow-twitch muscles are needed for what?

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2 years ago

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A 4 kg mass is in free fall. What is the velocity of the mass after 11 seconds

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3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

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3 years ago